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3 years ago

Which function is represented by the graph? f(x) = −2|x| + 1 f(x) = |x| + 1 f(x) = -2|x + 1| f(x) = |x + 1|

Mathematics

2 answers:

katrin [286]3 years ago

8 0

Answer:

its A

Step-by-step explanation:

I know things...

nevsk [136]3 years ago

7 0

The answer is A. F(x)= -2 |x| + 1

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Learning Theory In a typing class,the averege number N of words per minutes typed after t weeks of lessons can be modeled by N =

Jet001 [13]

Answer:

a) $N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

b) $N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

c) $70 =\frac{95}{1+8.5 e^{-0.12t}}$

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

d) If we find the limit when t tend to infinity for the function we have this:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

Step-by-step explanation:

For this case we have the following expression for the average number of words per minutes typed adter t weeks:

$N(t) = \frac{95}{1+8.5 e^{-0.12t}}$

Part a

For this case we just need to replace the value of t=10 in order to see what we got:

$N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

So the number of words per minute typed after 10 weeks are approximately 27.

Part b

For this case we just need to replace the value of t=20 in order to see what we got:

$N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

So the number of words per minute typed after 20 weeks are approximately 54.

Part c

For this case we want to solve the following equation:

$70 =\frac{95}{1+8.5 e^{-0.12t}}$

And we can rewrite this expression like this:

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

Now we can divide both sides by 8.5 and we got:

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

And then if we solve for t we got:

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

And we can see this on the plot 1 attached.

Part d

If we find the limit when t tend to infinity for the function we have this:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

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mars1129 [50]

Answer:

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Step-by-step explanation:

radius is half of what the diameter is.

50 / 2 = 25

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